Given an example of mapping $f:S\to S$ which is one-to-one but not onto.
Let's take $S=\mathbb{R}$ and $f(x)=e^x$ then it is clear that this function is injective but not onto.
Can anyone also give another example?
[Math] Function injective but not surjective
elementary-set-theory
Related Solutions
If the only thing you know about $X$ is that it is infinite, then it is not possible to describe such a mapping precisely enough that you can know what it does to each element. You don't even know what the elements are, so how would the example be able to tell what happens to them?
Using the Axiom of Choice, one can prove (see below) that every infinite $X$ has a subset $Y$ that is in bijective correspondence with $\mathbb N$. In that case you can construct $f$ as "for every element $x$ that corresponds to a natural number $n$, $f(x)$ is the element that corresponds to $n+1$; if $x$ doesn't correspond to a natural, $f(x)=x$.
Proof of above claim. Let $X$ be infinite; we want to construct a injection $\varphi:\mathbb N\to X$. Then the image of $\varphi$ will be the requested $Y$.
Fix a choice function $g:\mathcal P(X)\setminus\{\varnothing\}\to X$, and define $\varphi$ recursively: $$ \varphi(n) = g\bigl( X\setminus \{\varphi(i) \mid i<n\} \bigr) $$ The argument $X\setminus\{\varphi(i) \mid i<n\}$ is always non-empty because $X$ is infinite and there are only finitely many $\varphi(i)$ with $i<n$. So there are elements in $X$ that don't get removed.
Then $\varphi$ is injective: If $b>a$, then $\varphi(a)\ne\varphi(b)$ because $\varphi(b)$ will have been constructed as $g$ of a set where $\varphi(a)$ has explicitly been removed.
(Actually one needs only the axiom of Countable Choice, but that's a bit more involved -- see sketch by Asaf in the comments).
It may be that the downvotes are because your work does not even exhibit that you understand the concepts of injective and surjective, the definitions of which you may well be expected to use in your answers for b) and c).
Both of your answers are dead-wrong: the function listed in b) is NOT from $\Bbb N \to \Bbb N$ (it has the wrong co-domain). For example, $f(1) = \frac{1}{2}$ is NOT a natural number.
The function you give in c) IS surjective, but it also is injective, To see this, suppose:
$f(x) = f(y) \implies x - 1 = y - 1 \implies (x - 1) + 1 = (y - 1) + 1 \implies x = y$.
(EDIT: as pointed out in the comments, $f$ is not even a function from $\Bbb N \to \Bbb N$, as one can see by noting $f(0) = -1 \not\in \Bbb N$).
I suggest you try some function $f$ for b) that "skips" values in $\Bbb N$, you want "gaps" in the co-domain. For c), you might try using the floor function, somehow.
a) is the most important question, here though. Finiteness is key, that's what b) and c) are supposed to convince you of. Start by assuming $f$ is surjective. What happens if you assume (by way of contradiction), that $f$ is not injective? (hint: compare the cardinalities of the range, and the domain).
Best Answer
Take $S = \mathbb{N}$ and $f(n) = n+1$.